First slide

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Question

A thermodynamic system undergoes cyclic process ABCDA as shown in figure. The work done by the system in the cycle is

Easy

A

P₀V₀
B

2P₀V₀
C

P₀V₀/2
D

Zero

Solution

Area of the upper triangle
$= \frac{1}{2} \times \left( {2{V_0} - {V_0}} \right) \times \left( {3{P_0} - 2{P_0}} \right) = \frac{{{P_0}{V_0}}}{2}$
So work done in that cycle
${W_1} = \frac{{ - {P_0}{V_0}}}{2}$ [Anticlockwise]
Area of the lower triangle
$= \frac{1}{2} \times \left( {2{V_0} - {V_0}} \right) \times \left( {2{P_0} - {P_0}} \right) = \frac{{{P_0}{V_0}}}{2}$
So work done in this cycle
${W_2} = \frac{{ + {P_0}{V_0}}}{2}$ [Clockwise]
Total workdone W = W₁ + W₂ = 0

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